Problem: A few families took a trip to an amusement park together. Tickets cost $$8.50$ each for adults and $$3.00$ each for kids, and the group paid $$55.00$ in total. There were $3$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Answer: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${8.5x+3y = 55}$ ${x = y-3}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-3}$ for $x$ in the first equation. ${8.5}{(y-3)}{+ 3y = 55}$ Simplify and solve for $y$ $ 8.5y-25.5 + 3y = 55 $ $ 11.5y-25.5 = 55 $ $ 11.5y = 80.5 $ $ y = \dfrac{80.5}{11.5} $ ${y = 7}$ Now that you know ${y = 7}$ , plug it back into ${x = y-3}$ to find $x$ ${x = }{(7)}{ - 3}$ ${x = 4}$ You can also plug ${y = 7}$ into ${8.5x+3y = 55}$ and get the same answer for $x$ ${8.5x + 3}{(7)}{= 55}$ ${x = 4}$ There were $4$ adults and $7$ kids.